Let $\tau_1$ and $\tau_2$ be two topological spaces on a set $X$. Then $\tau_1$ is said to be a finer topology than $\tau_2$ if $\tau_1\supset\tau_2$.
Prove that the Euclidean topology in $\mathbb{R}$ is finer than the finite-closed topology.
Let's define the cofinite topology$\tau_c=\{A:\mathbb{R}\setminus A\: \text{is finite} \}$
Therefore $A$ cannot be an open interval of the form $(a,b)$ once the compliment is not finite. $A$ can be an interval of the form $(-\infty,a]\cup[b,+\infty)$ or the form $(-\infty,x)\cup (a,+\infty)$. So both of them are contained in a open set in the usual topology on $\mathbb{R}$, generated by the open sets, which proves the assertion.
Best Answer
If $F$ is a finite subset of $\mathbb{R}$, enumerate it as $x_1 < x_2 < x_3 < \ldots x_n$. Then $$X\setminus F = (-\infty, x_1) \cup (x_1, x_2) \cup (x_2, x_3) \cup \ldots \cup (x_n, +\infty)$$
which is a union of open intervals/segments so usual open.
So all sets in $\tau_c$ (these are of the form $X\setminus F$ with $F$ finite, or equal to $\emptyset$) are "usual open", and this is what you had to show.